A 71.5-kg diver is standing at the end of a diving board while it is vibrating u

Question

A 71.5-kg diver is standing at the end of a diving board while it is vibrating up and down in simple harmonic motion, as indicated in the figure. The diving board has an effective spring constant of k = 3060 N/m, and the vertical distance between the highest and lowest points in the motion is 0.201 m. (a) What is the amplitude of the motion? (b) Starting when the diver is at the highest point, what is his speed one-quarter of a period later? (c) If the vertical distance between his highest and lowest points were changed to 0.135 m, what would be the time required for the diver to make one complete motional cycle?

Explanation / Answer

k = 3060 m

m = 71.5 kg

angular frequency is given as

w = sqrt(k/m) = sqrt (3060/71.5) = 6.54 rad/s

d = distance between highest and lowest point = 0.201 m

a)

A = distance of midpoint from highest or lowest point = d/2 = 0.201/2 = 0.1005 m

b)

in quarter time , diver reach the equilibrium position, hence speed is given as

V = AW = 0.1005 x 6.54 = 0.66 m/s

c)

Time period is independent of distance between lowest and highest points, so T remains same as

T = 2pi/w = 2 x 3.14 /6.54 = 0.96 sec

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